= Solution
Freeze a uniform state with $0<\alpha<1$, $\rho_i>0$, and seek a finite <characteristic speed> $c$. Denote its perturbation amplitudes by $a,v_1,v_2,\pi$. The two <phase>-mass equations and two momentum equations give
$$
(u_1-c)a+\alpha v_1=0,\qquad-(u_2-c)a+(1-\alpha)v_2=0,
$$
$$
\rho_1(u_1-c)v_1+\pi=0,\qquad\rho_2(u_2-c)v_2+\pi=0.
$$
Eliminating the amplitudes gives
$$
\rho_1(1-\alpha)(u_1-c)^2+\rho_2\alpha(u_2-c)^2=0.
$$
With $s^2=\rho_1(1-\alpha)/(\rho_2\alpha)>0$, factor the quadratic over the complex numbers to obtain
$$
\boxed{c_\pm=\frac{u_2\pm is u_1}{1\pm is}=\frac{u_2+s^2u_1}{1+s^2}\pm i\frac{s(u_1-u_2)}{1+s^2}.}
$$
For nonzero relative <velocity> the speeds are a nonreal conjugate pair. A <Fourier mode>[Fourier perturbation] $e^{ik(x-ct)}$ on the growing branch has magnitude $e^{|k|\,|\operatorname{Im}c|t}$. Arbitrarily short waves therefore grow arbitrarily quickly, preventing a continuous dependence estimate in the usual finite-order <Sobolev norms>. This generic slip state is <ill-posed> as an <initial value problem>.
The <pressure> has no independent time-evolution equation; it enforces the incompressible <volume> <constraint>, so the request concerns the finite <characteristic speeds> rather than a freely propagating <pressure> mode. At $u_1=u_2$ the two speeds coincide and this particular nonreal-speed argument disappears; further degeneracy analysis is needed there, rather than calling coincident real speeds complex. The issue is <complex characteristic speeds of a pressure-coupled two-fluid model>.
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